Hint

Answer

9408!
The rule of divisibility says that a number is a multiple of 4 if its last 2 numbers are multiples of 4. So the last 2 numbers can be 21 different options:00, 04, 08, 12, 16, 20, 24, 28, 40, 44, 48, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96 because they are multiples of 4 and do not have 3's or 5's.
We have the last 2 digits we need to know how many can be in the first digit: 1, 2, 4, 6, 7, 8, 9 and the second and third numbers can have these 7 and number 0 too.

By using permutations we give the possible numbers in each digit and the last 2 digits together.
1st*2nd*3rd*(4th and 5th)=?
7x8x8x21=9408Hide